Chemistry, 17.07.2019 05:00, soulspiritsa

For the chemical reaction hclo4(aq)+naoh(aq)⟶h2o(l)+naclo4(aq ) write the net ionic equation, including the phases.

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The net ionic equation is as below

H^+(aq) + H^-(aq) → H2O(l)

Explanation
from the chemical equation

HClO4(aq) + NaOH (aq) → H2O(l) + NaClO4(aq)

write ionic equation
that is,
H^+(aq)+ClO4^-(aq) + Na^+(aq) +OH^-(aq)→ H2O(l) + Na^+(aq)+ClO4^-(aq)

cancel the spectator ions in both side(ions that do no take place in a chemical equation)
that ClO4^- and Na^+

the net ionic is therefore
= H^+(aq) + OH^-(aq) → H2O(l)

answered: 19wawrzkeek

The net ionic equation is defined as the removing spectator ions from total equation and list the species that only participate in the reaction. In the given equation, $\text {HClO}_{4}_{(aq)}+\text{NaOH}_{(aq)}\rightarrow\text{H}_{2}\text O_{(l)}+\text {NaClO}_{4}_{(aq)}$ , the net iconic equation will be:

$H^{+}_{(aq)} + \text{OH}^{-}_{(aq)}\rightarrow \text{H}_{2}\text{O}_{(l)}$

In aqueous solution, HClO $_{4}$ will dissolve into H $^{+}$ and ClO $_{4}^{-}$ . Similarly, NaOH will be dissociated into Na $^{+}$ and OH-. The resultant salt formed is NaClO $_{4}$ . The equation can be expanded as:

$\text {HClO}_{4}_{(aq)}+\text{NaOH}_{(aq)}\rightarrow\text{H}_{2}\text O_{(l)}+\text {NaClO}_{4}_{(aq)}\\\\\text {H}^{+} + \text {ClO}_{4}^{-} + \text {Na}^{+} + \text {OH}^{-}\rightarrow \text{H}_{2}\text {O} + \text{Na}^{+}+ \text {ClO}_{4}^{-}$

Thus, cancelling out the terms on both sides of the equation, those are termed as spectator ions. The spectator ions are:

$\text {H}^{+} + \text {ClO}_{4}^{-} + \text {Na}^{+} + \text {OH}^{-}\rightarrow \text{H}_{2}\text {O} + \text{Na}^{+}+ \text {ClO}_{4}^{-}$

Remaining, net iconic equation is:

$H^{+}_{(aq)} + \text{OH}^{-}_{(aq)}\rightarrow \text{H}_{2}\text{O}_{(l)}$

Therefore, for any strong acid and strong basic reaction, the net ionic reaction is the neutralization of Hydrogen and Hydroxide ions to form water.

To know more net iconic equations, refer to the following link:

link

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

sure im new too!

explanation: